Solving for cos θ gives us:

`cos theta=+-1/4`

If `cos alpha=1/4`, then the reference angle is α = 1.3181.

So for `cos theta=1/4`, we have θ in the first and 4th quadrants. So

`θ = 1.3181 or 4.9651`

For `cos theta=-1/4`, we have θ in the 2nd and 3rd quadrants. So

`θ = 1.8235 or 4.4597`

So ` θ= 1.3181, 1.8235,` ` 4.4597` ` or 4.9651` radians.

We can see from the graph of `y=cos^2theta-1/16` that our answer is correct:

graph of y=cos^2theta-1/16